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Calculation when Mixing Two Compounds...


New member
Sep 4, 2013
Dear Comunity,

I am running into a (probably pretty simple) problem here, yet the most simple things are sometimes upon me...:(

I have a liquid solution which consists of 70% compound A and 30% compound B.
I also have a bottle with 100% compound A.

What I would like to do is to achieve a concentration of 80% compound A and 20% compound B. Can someone tell me how this is calculated?

I was thinking of something like c1*v1=c2*v2 but this doesnt get me anywhere....am I missing something?

Any help on this would be very much appreciated!

Best regards,


Staff member
Feb 24, 2012
I would set it up as follows:

Let $x$ be the amount of solution you currently have, the 7:3 mix. Then let $y$ be the amount of 100% compound A you wish to add. Thus we want:

\(\displaystyle 0.7x+y=0.8(x+y)\)

Do you see how both sides of the equation represent the amount of compound A present in the final solution? The left side represents the initial amount of compound A plus the amount added. The right side reflects the fact that we want 80% of the final amount to be compound A.

Now you may solve this for $y$ as a function of $x$. What do you find?


Well-known member
Feb 2, 2012
Hello, haubenkoch!

We can disregard references to compound B.

I have a liquid solution which is 70% alcohol.
I also have a bottle with 100% alcohol.

I would like a concentration of 80% alcohol.

This is an explanation of MarkFL's solution.

Let [tex]x[/tex] = liters of 70% alcohol.
It contains [tex]0.7x[/tex] liters of alcohol.

We add [tex]y[/tex] liters of 100% alcohol.
It contains [tex]y[/tex] liters of alcohol.

The mixture contains [tex]0.7x + y[/tex] liters of alcohol. .[1]

But we know that the mixture will be
. . [tex]x+y[/tex] liters which is 80% alcohol.
It contains [tex]0.8(x+y)[/tex] liters of alcohol. .[2]

We just described the final amount of alcohol in two ways.

There is our equation! ..[tex]0.7x + y \:=\:0.8(x+y)[/tex]